Question: Is ${536430}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {536430}= &&{5}\cdot100000+ \\&&{3}\cdot10000+ \\&&{6}\cdot1000+ \\&&{4}\cdot100+ \\&&{3}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {536430}= &&{5}(99999+1)+ \\&&{3}(9999+1)+ \\&&{6}(999+1)+ \\&&{4}(99+1)+ \\&&{3}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {536430}= &&\gray{5\cdot99999}+ \\&&\gray{3\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {5}+{3}+{6}+{4}+{3}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${536430}$ is divisible by $3$ if ${ 5}+{3}+{6}+{4}+{3}+{0}$ is divisible by $3$ Add the digits of ${536430}$ $ {5}+{3}+{6}+{4}+{3}+{0} = {21} $ If ${21}$ is divisible by $3$ , then ${536430}$ must also be divisible by $3$ ${21}$ is divisible by $3$, therefore ${536430}$ must also be divisible by $3$.